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Angular momentum in QM

  1. Nov 14, 2006 #1
    Hello,

    I'm trying to solve a problem dealing with finding the probability of measuring certain values of [tex]L^2[/tex] for a particle.

    The particle is on a sphere and is in the state [tex]\Psi (\theta , \phi) = Ne^{\\cos{\theta }}[/tex].

    I don't know quite how to start, I guess I have to decompose the wave function in eigenfunctions for [tex]L^2[/tex], and then find the corresponding eigenvalues, and form that find the probability of measuring that particular eigenvalue, but like I said, I don't really know where to start.

    Does anybody have any pointers?

    Thanks!
     
    Last edited: Nov 14, 2006
  2. jcsd
  3. Nov 15, 2006 #2

    dextercioby

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    Nope, you have to do you wrote. That is try to write your wavefunction as a linear combination of angular momentum eigenstates.

    Daniel.
     
  4. Nov 15, 2006 #3
    Thanks for the reply!

    I've worked some more, and I think I've made some progress :smile:

    The eigenfunctions of [tex]\^{L}^2[/tex] are spherical harmonics, but since the wave function is independent of [tex]\phi[/tex], m = 0, i.e [tex]\Psi(\theta,\phi) = \sum_{l} c_{l,0}Y_l^0 [/tex].
    The eigenvalues are [tex]l(l+1)\hbar[/tex]. So I just see which value of [tex]l[/tex] corresponds with the values I'm supposed to find the probability of. The probability is calculated with: [tex]|c_{l}|^2 = |<\Psi|Y_l^0>|^2[/tex]. This is where I'm stuck at the moment. The integrals are really complicated and I can't find them in any books, is there any easier way to calculate this?

    Thanks!
     
    Last edited: Nov 15, 2006
  5. Nov 15, 2006 #4

    nrqed

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    A suggestion: I don't recall the exact form of the[itex] Y^l_0 [/itex], but aren't they simple to write in terms of [itex] (cos \theta)^n [/itex] ? Then you simply have to Taylor expand [itex] e^{cos \theta)} [/itex]. The coefficients will simply be the usual 1/n!. If you have a closed form expression for the [itex] Y^l_0 (cos (\theta) [/itex], then you can find the coeeficients c_l without doing a single integral.

    Patrick
     
  6. Nov 15, 2006 #5
    That probably would have worked, but I managed to solve the damn integrals before I read your reply :smile: All it took was some (a lot of) integration by parts, and it turned out ok. Now I can finally rest. :smile:

    Thanks for taking the time guys!
     
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